What is the integral of (x+16)/(x^2+2x-8) ?

The integral of (x+16)/(x^2+2x-8) is 1/2 ln|x^2+2x-8|-5(1/2 ln| 1/3 (x+1)+1|-1/2 ln| 1/3 (x+1)-1|)+C

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integral of (x+16)/(x^2+2x-8)

\int \frac{x + 16}{x ^{2} + 2 x - 8} d x

\frac{1}{2} ln x^{2} + 2 x - 8 - 5 (\frac{1}{2} ln \frac{1}{3} (x + 1) + 1 - \frac{1}{2} ln \frac{1}{3} (x + 1) - 1) + C

Solution steps

\int \frac{x + 16}{x ^{2} + 2 x - 8} d x

Complete the square

x^{2} + 2 x - 8 : (x + 1)^{2} - 9

= \int \frac{x + 16}{( x + 1 ) ^{2} - 9} d x

Apply u-substitution

: \int \frac{u + 15}{u ^{2} - 9} d u

= \int \frac{u + 15}{u ^{2} - 9} d u

Expand

\frac{u + 15}{u ^{2} - 9} : \frac{u}{u ^{2} - 9} + \frac{15}{u ^{2} - 9}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{u}{u ^{2} - 9} d u + \int \frac{15}{u ^{2} - 9} d u

\int \frac{u}{u ^{2} - 9} d u = \frac{1}{2} ln u^{2} - 9

\int \frac{15}{u ^{2} - 9} d u = - 5 (\frac{1}{2} ln \frac{u}{3} + 1 - \frac{1}{2} ln \frac{u}{3} - 1)

= \frac{1}{2} ln u^{2} - 9 - 5 (\frac{1}{2} ln \frac{u}{3} + 1 - \frac{1}{2} ln \frac{u}{3} - 1)

Substitute back

u = x + 1

= \frac{1}{2} ln (x + 1)^{2} - 9 - 5 (\frac{1}{2} ln \frac{x + 1}{3} + 1 - \frac{1}{2} ln \frac{x + 1}{3} - 1)

Simplify

\frac{1}{2} ln (x + 1)^{2} - 9 - 5 (\frac{1}{2} ln \frac{x + 1}{3} + 1 - \frac{1}{2} ln \frac{x + 1}{3} - 1) : \frac{1}{2} ln x^{2} + 2 x - 8 - 5 (\frac{1}{2} ln \frac{1}{3} (x + 1) + 1 - \frac{1}{2} ln \frac{1}{3} (x + 1) - 1)

= \frac{1}{2} ln x^{2} + 2 x - 8 - 5 (\frac{1}{2} ln \frac{1}{3} (x + 1) + 1 - \frac{1}{2} ln \frac{1}{3} (x + 1) - 1)

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 2 x - 8 - 5 (\frac{1}{2} ln \frac{1}{3} (x + 1) + 1 - \frac{1}{2} ln \frac{1}{3} (x + 1) - 1) + C

x \to 0 lim (x^{6} cos (\frac{4}{x}))

\int_{0}^{\frac{π}{4}} 7 tan^{3} (x) d x

\int \frac{3 x}{( x ^{2} + 3 ) ^{2}} d x

f (x) = x + 2

x \to - 1 lim (x^{2} - 3 x + 1)

What is the integral of (x+16)/(x^2+2x-8) ?
The integral of (x+16)/(x^2+2x-8) is 1/2 ln|x^2+2x-8|-5(1/2 ln| 1/3 (x+1)+1|-1/2 ln| 1/3 (x+1)-1|)+C